7. Product and Process Comparisons
7.4. Comparisons based on data from more than two processes
7.4.2. Are the means equal?

## Assessing the response from any factor combination

Estimating factor level means

Factor level mean example for a 4 level treatment (or 4 different treatments)

Definition of contrasts and orthogonal contrasts

Estimation of contrasts

Confidence intervals for contrasts

Contrast example

Estimasting other linear combinations

This page treats how to estimate and put confidence bounds around the response to different combinations of factors.  Primary focus is on the combinations that are known as contrasts. We begin, however, with the simple case of a single factor level mean.

Estimation of a Factor Level Mean With Confidence Bounds

An unbiased estimate of the factor level mean mi in the 1-way ANOVA model is given by:

where
The variance of this sample mean estimate is

It can be shown that:
is distributed as t-distribution statistic with (N- k) degrees of freedom for the ANOVA model under consideration, where N is the total number of observations and k is the number of factor levels or groups. The degrees of freedom are the same as were used to calculate the MSE in the ANOVA table. That is: dfe (degrees of freedom for error) = N - k. From this we can calculate 1-a confidence limits for each µi. These are, given by:
Example 1

The data in the accompanying table resulted from an experiment run in a completely randomized design in which each of four treatments was replicated five times.
 Total Mean Group 1 6.9 5.4 5.8 4.6 4.0 26.70 5.34 Group 2 8.3 6.8 7.8 9.2 6.5 38.60 7.72 Group 3 8.0 10.5 8.1 6.9 9.3 42.80 8.56 Group 4 5.8 3.8 6.1 5.6 6.2 27.50 5.50 All Groups 135.60 6.78

This experiment can be illustrated by the table layout for this 1-way ANOVA experiment shown below:

The resulting ANOVA table is

The estimate for the mean of group 1 is 5.34, and the sample size is n1 = 5.

Since the confidence interval is two-sided, the entry a/2 value for the t-table is .5(1 - .95) = .025, and the associated degrees of freedom is N - 4,  or 20 - 4 = 16.

From t-tables or a computer program we obtain t.025;16 = 2.120.

Next we need the standard error of the mean for group 1:

Hence, we obtain confidence limits 5.34 ± 2.120 (0.5159) and the confidence interval is

Definition and Estimation of Contrasts

Definitions

A contrast is a linear combination of 2 or more factor level means with coefficients that sum to zero.

Two contrasts are orthogonal if the sum of the products of corresponding coefficients (i.e. coefficients for the same means) adds to zero.

Formally, the definition of contrast is expressed below, using the notation mi for the i-th treatment mean:

Simple contrasts includes the case of the difference between two factor means, such as m1 - m2 . If one wishes to compare treatments 1 and 2 with treatment 3, one way of expressing this is by: m1 + m2 - 2m3. Note that
m1 - m2 has coefficients +1, -1
m1 + m2 - 2m has coefficients +1, +1, -2.
These coefficients sum to zero.
As an example of orthogonal contrasts note the three contrasts defined by the table below, where the rows denote coefficients for the column treatment means.
The following is true:
1. The sum of the coefficients for each contrast is zero.

2. The sum of the products of coefficients of each pair of contrasts  is also 0 (orthogonality property).

3. The first two contrasts are simply pair wise comparisons, the third one involves all the treatments.

As might be expected, contrasts are estimated by taking the same linear combination of treatment mean estimates. In other words:

and
Note: These formulas hold for any linear combination of treatment means, not just for contrasts.

Confidence Interval for a Contrast

An unbiased estimator for a contrast C is given by

The estimate of  is

The estimator  is normally distributed because it is a linear combination of independent normal random variables. It can be shown that:

for the one way ANOVA model under discussion.

Therefore, the 1-a confidence limits for C are:

Example 2

We wish to estimate, in our previous example, the following contrast:

and construct a 95 percent confidence interval for C.

The point estimate is:

For a confidence coefficient of 95 % and df = 20 - 4 = 16,
t.025;16 = 2.12. Therefore, the desired 95 % confidence interval is  -.5 ± 2.12(.5158) or

(-1.594, 0.594).
Estimation of Linear Combinations

Sometimes we are interested in a linear combination of the factor level means that is not a contrast. Assume that in our sample experiment certain costs are associated with each group. For example there might be costs associated with each factor as follows:

Factor      Cost in \$
1               3
2               5
3               2
4               1

The following linear combination might then be of interest:

This resembles a contrast, but the coefficients ci do not sum up to zero. A linear combination has definition:
with no restrictions on the coefficients ci.

Confidence limits for a linear combination C are obtained in precisely the same way as those for a contrast, using the same calculation for the point estimator and estimated variance.